I know it has pole of order 2 at z = 0 but I'm not sure it has essential singularity at z = $\infty$.
In $C_{\infty}$ , $\pm\infty , \pm i\infty$ are same point. They should give same value if it doesn't have essential singularity at z = $\infty$.
z = $\infty$ , $lim_{w\to0}\frac{1}{\frac{1}{w}(e^{\frac{1}{w}}-1)} = 0$
but z = $i\infty$ , $lim_{w\to0}\frac{1}{\frac{i}{w}(e^{\frac{i}{w}}-1)} = lim_{w\to0}\frac{1}{\frac{i}{w}(\cos{\frac{1}{w}}+i\sin{\frac{1}{w}}-1)}$
I think it isn't analytic at z = $i\infty$.
In the context of complex analysis, there is one point at infinity, denoted by $\infty$. The function $f$ has a singularity at $\infty$ if $f\left(\frac1z\right)$ has a singularity at $0$. But\begin{align}f\left(\frac1z\right)&=\frac1{\frac1z\left(e^{\frac1z}-1\right)}\\&=\frac z{e^{\frac1z}-1}\end{align}Note that this is undefined when $z$ has the form $\frac1{2\pi i}$, for some $n\in\mathbb{Z}\setminus\{0\}$. Therefore, if you are talking of isolated singularities, then $\infty$ isn't one. Otherwise, it is a limit point of isolated singularities.